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Solution: A 2.950 × 10−2 M solution of NaCl in water is at 20.0°C. The sample was created by dissolving a sample of NaCl in water and then bringing the volume up to 1.000 L. It was determined that the volume of

Problem

A 2.950 × 10−2 M solution of NaCl in water is at 20.0°C. The sample was created by dissolving a sample of NaCl in water and then bringing the volume up to 1.000 L. It was determined that the volume of water needed to do this was 999.2 mL . The density of water at 20.0°C is 0.9982 g/mL.

Part A. Calculate the molality of the salt solution. Express your answer to four significant figures and include the appropriate units.

NaCl =

Part B. Calculate the mole fraction of salt in this solution. Express the mole fraction to four significant figures.

χ NaCl =

Part C. Calculate the concentration of the salt solution in percent by mass. Express your answer to four significant figures and include the appropriate units.

percent by mass NaCl =

Part D. Calculate the concentration of the salt solution in parts per million. Express your answer as an integer to four significant figures and include the appropriate units.

parts per million NaCl =

Solution

Based on the information provided to prepare a 2.950x10-2 M NaCl solution in water at 20 °C, we´re asked to calculate the concentration of the salt solution in 4 parts

Molality (m), mole fraction (X), percent by mass (%), and parts per million (ppm).

For all cases, we need the amount of NaCl in the sample.

We calculate the moles of NaCl from the definition of Molarity (M):

$\overline{){\mathbf{Molarity}}{\mathbf{\left(}}{\mathbf{M}}{\mathbf{\right)}}{\mathbf{=}}\frac{\mathbf{moles}}{\mathbf{Liters}}}$

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